Physics 107 HOMEWORK ASSIGNMENT #16
Cutnell & Johnson, 7th edition
Chapter 19: Problem 60
Chapter 20: Problems 20, 26, 44, 123
**60 A positive charge of +q1 is located 3.00 m to the left of a negative charge −q2. The charges
have different magnitudes. On the line through the charges, the net electric field is zero at a spot
1.00 m to the right of the negative charge. On this line there are also two spots where the
potential is zero. Locate these two spots relative to the negative charge.
**20 A digital thermometer employs a thermistor as the temperature-sensing element. A
thermistor is a kind of semiconductor and has a large negative temperature coefficient of
resistivity . Suppose = –0.060 (C˚)-1 for the thermistor in a digital thermometer used to measure
the temperature of a patient. The resistance of the thermistor decreases to 85% of its value at the
normal body temperature of 37.0°C. What is the patient’s temperature?
*26 A piece of Nichrome wire has a radius of 6.5 x 10-4 m. It is used in a laboratory to make a
heater that uses 4.00 x 102 W of power when connected to a voltage source of 120 V. Ignoring
the effect of temperature on resistance, estimate the necessary length of wire.
*44 Multiple-Concept Example 9 reviews the concepts that are important to this problem. A
light bulb is wired in series with a 144 Ω resistor, and they are connected across a 120.0-V
source. The power delivered to the light bulb is 23.4 W. What is the resistance of the light bulb?
Note that there are two possible answers.
123 Concept Questions The drawing shows two circuits, and the same
battery is used in each. The two resistances RA in circuit A are the same, and
the two resistances RB in circuit B are the same. (a) How is the total power
delivered by the battery related to the equivalent resistance connected
between the battery terminals and to the battery voltage? (b) When two
resistors are connected in series, is the equivalent resistance of the
combination greater than, smaller than, or equal to the resistance of either
resistor alone? (c) When two resistors are connected in parallel, is the
equivalent resistance of the combination greater than, smaller than, or equal
to the resistance of either resistor alone? (d) The same total power is
delivered by the battery in circuits A and B. Is RB greater than, smaller than,
or equal to RA?
Problem Knowing that the same total power is delivered in each case, find
the ratio RB/RA for the circuits in the drawing. Verify that your answer is
consistent with your answer to Concept Question (d).
60. REASONING AND SOLUTION The information about the electric field requires that
k q2 /(1.00 m)2 = k q1 /(4.00 m)2
so
q2 = (1/16.0) q1
Since q2 is negative and q1 is positive, this result implies that
 q2  q1 /16.0
(1)
Let x be the distance of the zero-potential point from the negative charge and d be the
separation between the charges. Then, the total potential is
kq1/(d + x) + kq2/x = 0
if the point is to the right of q2 and
kq1/(d  x) + kq2/x = 0
if the point is between the charges and to the left of q2. Using Equation (1) and solving for x
in each case gives
x  0.200 m to the right of the negative charge
x  0.176 m to the left of the negative charge
20. REASONING AND SOLUTION The resistance of the thermistor decreases by 15%
relative to its normal value of 37.0 °C. That is,

R

RR
 0

0
.1
5
R
R
0
0
According to Equation 20.5, we have
R = R0[1 +  (T – T0)]
or (R – R0) =  R0(T – T0)
or
R

R
0


T

T

0
.
1
5


0
R
0
Rearranging this result gives


0
.
1
5

0
.
1
5
T

T

3
7
.
0

C


3
9
.
5
°
C
0

1



0
.
0
6
0
C

26. REASONING AND SOLUTION We know that the resistance of the wire can be obtained
from
P = V 2 /R
or
R = V 2 /P
We also know that R = L/A. Solving for the length, noting that A =  r2, and using
 = 100  10–8 .m from Table 20.1, we find




  






2
2 22
–
4
2
V
/
P
r
1
2
0
V
.
5

1
0
m
6
R
A
V
r
L




5
0
m
–
8
2
P
1
0
0

1
0


m
4
.
0
0

1
0
W
2
44. REASONING The circuit containing the light bulb and resistor is shown in the drawing.
2
The resistance R of the light bulb is related to the power delivered to it by R1  P1 / I
1
(Equation 20.6b), where I is the current in the circuit. The power is known, and the current
can be obtained from Ohm’s law as the voltage V of the source divided by the equivalent
resistance R of the series circuit: I V / RS. Since the two resistors are wired in series, the
S
equivalent resistance is the sum of the resistances, or RS R
1R
2.
Light bulb
R2 = 144 Ω
R1
P1 = 23.4 W
120.0 V
SOLUTION The resistance of the light bulb is
R1 
P1
I2
(20.6b)
Substituting I V / RS (Equation 20.2) into Equation 20.6b gives
2
P
R
P
P
1S
1
1
R



1 2
2
I 
V
/R
2 V
S
(1)
The equivalent resistance of the two resistors wired in series is RS R
1R
2 (Equation
20.16). Substituting this expression for RS into Equation (1) yields
2 P
2
P
R
RR

1
1
S 1
R


1
2
2
V
V
2
Algebraically rearranging this equation, we find that
2

V
2
2
R

2
R

R

R

0


1 2
1
2

P
1


This is a quadratic equation in the variable R1. The solution can be found by using the
quadratic formula (see Appendix C.4):
2

V

2
R


 2 P

1

R
1
2
2

V
2
2
R

4
R

2
 2 P

1

2
2


1
2
1
2
2
0
.
0
V
2
0
.
0
V





2

2
1
4
4
2
1
4
4
4
1
4
4

 

2
3
.4W
2
3
.4W





2
8
5
.9a
n
d2
4
2
123. CONCEPT QUESTIONS
a. The total power P delivered by the battery is related to the equivalent resistance Req
connected between the battery terminals and to the battery voltage V according to
2
Equation 20.6c: PV / Req.
b. When two resistors are connected in series, the equivalent resistance RS of the
combination is greater than the resistance of either resistor alone. This can be seen directly
from RS R
1R
2 (Equation 20.16).
c. When two resistors are connected in parallel, the equivalent resistance RP of the
combination is smaller than the resistance of either resistor alone. This can be seen directly

1

1

1
by substituting values in R
P R
1 R
2 (Equation 20.17) or by reviewing the discussion
in Section 20.7 concerning the water flow analogy for electric current in a circuit.
d. Since the total power delivered by the
2
battery is PV / Req and since the power
and the battery voltage are the same in both
cases, it follows that the equivalent
resistances are also the same. But the
parallel combination has an equivalent
resistance RP that is smaller than RB,
whereas the series combination has an
equivalent resistance RS that is greater than
RA. This means that RB must be greater
than RA, as Diagram 1 at the right shows. If
RA were greater than RB, as in Diagram 2,
the equivalent resistances RS and RP would
not be equal.
Resistance
Resistance
RS
Diagram 1
RB
RA
RS = RP
RB
RA
RP
Diagram 2
SOLUTION As discussed in our answer to Concept Question (d), the equivalent
resistances in circuits A and B are equal. According to Equations 20.16 and 20.17, the
series and parallel equivalent resistances are
R
R
R
2
R
S
A
A
A
1 1 1
1
 
o
r R
P2R
B
R
R
R
P
B
B
Setting the equivalent resistances equal gives
1
2
R

Ro
r
A
2B
As expected, RB is greater than RA.
R
B

4
R
A